Every Separable Complex Fréchet Space with a Continuous Norm Is Isomorphic to a Space of Holomorphic Functions

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Every Separable Complex Fréchet Space with a Continuous Norm Is Isomorphic to a Space of Holomorphic Functions Every separable complex Fr´echet space with a continuous norm is isomorphic to a space of holomorphic functions Jos´eBonet Abstract Extending a result of Mashreghi and Ransford, we prove that every complex sep- arable infinite dimensional Fr´echet space with a continuous norm is isomorphic to a space continuously included in a space of holomorphic functions on the unit disc or the complex plane, which contains the polynomials as a dense subspace. As a con- sequence examples of nuclear Fr´echet spaces of holomorphic functions without the bounded approximation exist. 1 Introduction. Let G be an open connected domain in the complex plane C. We denote by H(G) the Fr´echet space of holomorphic functions on G, endowed with the topology of uniform conver- gence on compact subsets of G. A Fr´echet space E of holomorphic functions on the domain G is a Fr´echet space that is a subset of H(G), such that the inclusion map E ⊂ H(G) is continuous and E contains the polynomials. By the closed graph theorem, the inclusion map E ⊂ H(G) is continuous if and only if the point evaluations at the points of G are continuous on E. If the polynomials are dense in E, then E is separable. The set P of polynomials is dense in H(G) if and only if G is the whole complex plane or a simply connected domain; see [15, Theorem 13.11]. However, this assumption is not needed in our results below. Banach spaces of holomorphic functions on the unit disc D and on the complex plane C have been thoroughly investigated. We refer the reader for example to the books [8], [19] and [20]. H¨ormander Fr´echet algebras of entire functions [2], [3], [12], the Fourier-Laplace transform of spaces of (ultra)-distributions [7] and intersections of growth Banach spaces [5], [8] are natural examples of Fr´echet spaces of holomorphic func- arXiv:2002.12677v1 [math.FA] 28 Feb 2020 tions. Vogt [18] proved that there are Fr´echet spaces E which are contained in H(G) such that the inclusion E ⊂ H(G) is not continuous. These examples are not Fr´echet spaces of holomorphic functions on G in our sense. Very recently, Mashreghi and Ransford [11, The- orem 1.3] have shown that every separable, infinite-dimensional, complex Banach space Y is isometrically isomorphic to a Banach space of holomorphic functions X ⊂ H(D) such that the polynomials are dense in X. The purpose of this note is to extend this result to the setting of Fr´echet spaces and to derive a few consequences. Our notation for functional analysis, in particular for Fr´echet spaces and their duals, is standard. We refer the reader to [9], [10], [13] and [14]. If E is a Fr´echet space, its topological dual is denoted by E′. The weak topology on E is denoted by σ(E, E′) and the weak* topology on E′ by σ(E′, E). The linear span of a subset A of E is denoted by span(A). In what follows, we set N0 := N ∪ {0}. 2020 Mathematics Subject Classification. Primary: 46A04, secondary: 46A11, 46A32. Key words and phrases. Spaces of holomorphic functions, Fr´echet spaces, continuous norm, bounded approximation property 1 2 Results. We start with the following observation. If a Fr´echet space F is isomorphic to a Fr´echet space E ⊂ H(G) of holomorphic functions on an open connected domain G ⊂ C, then F has a continuous norm. Indeed, let T : F → E be a topological isomorphism and let K ⊂ G be an infinite compact set. Since both T and the inclusion E ⊂ H(G) are continuous by assumption, there is a seminorm p on F such that supz∈K |T (x)(z)|≤ p(x) for each x ∈ F . If p(x) = 0 for some x ∈ F , then T (x)(z) = 0 for each z ∈ K. Therefore, the holomorphic function T (x) vanishes on G. Since T is injective, x = 0. We need the following Lemma to prove our main result. Lemma 2.1 Let F be a separable, infinite dimensional Fr´echet space with a continuous ′ ′ norm. Then one can find two sequences (en)n∈N0 ⊂ F and (en)n∈N0 ⊂ F such that (i) span{en | n ∈ N0} is dense in F . ′ ′ (ii) (en)n∈N0 is equicontinuous in F . ′ ′ ′ (iii) span{en | n ∈ N0} is dense in (F ,σ(F F )). ′ ′ N ′ (iv) hen, emi = 0 if n 6= m and hen, eni = 1 for each n ∈ 0. That is, (en, en)n∈N0 is a biorthogonal system. Proof. We denote by τ the metrizable topology of the Fr´echet space F and by p the continuous norm on F . Let U := {x ∈ F | p(x) ≤ 1} be the unit ball of the norm p. The polar set U ◦ of U in F ′ coincides with {u ∈ F ′ | |u(x)| ≤ p(x) for all x ∈ F } and, by ◦ ◦ ′ ′ Hahn-Banach theorem, its linear span H := span(U )= ∪s∈NsU is σ(F , F )-dense in F . Since (F,τ) is separable, there is a countable, infinite subset A = (yn)n∈N0 of F which is τ dense in F . Hence, A is also dense in the normed space (F,p), and this normed space is separable, too. The topological dual of (F,p) coincides with H = span(U ◦) ⊂ F ′. Since (F,p) is metrizable and separable, it follows from [14, Corollary 2.5.13] that H is σ(H, F )- ′ separable. We select a countable, infinite subset A = (vn)n∈N0 of H which is dense in (H,σ(H, F )). We now apply [14, Proposition 2.3.2] to find sequences (xn)n∈N0 ⊂ span(A) ′ ′ ′ ′ and (xn)n∈N0 ⊂ span(A ) such that hxn,xmi = 0 if n 6= m and hxn,xni = 1 for each N N ′ N ′ n ∈ 0, and span({xn,n ∈ 0}) = span(A) and span({xn,n ∈ 0}) = span(A ). N ′ ′ ◦ N Since for each n ∈ 0 we have xn ∈ span(A ) ⊂ H = ∪s∈NsU , then for each n ∈ 0 we ′ ′ −1 ′ can select m(n) > 0 such that |xn(x)|≤ m(n)p(x) for each x ∈ F . We set en := m(n) xn and en := m(n)xn for each n ∈ N0 and we check that these sequences satisfy properties (i) − (iv). (i) span({en,n ∈ N0}) = span({xn,n ∈ N0}) = span(A) is dense in (F,τ). ′ N ′ (ii) |en(x)|≤ p(x) for each x ∈ F and each n ∈ 0. Therefore, (en)n∈N0 is equicontin- uous in F ′. ′ N ′ N ′ (iii) span({en,n ∈ 0}) = span({xn,n ∈ 0}) = span(A ) is σ(H, F )-dense in H. Since ′ ′ N ′ ′ H is σ(F , F )-dense in F , we conclude that span{en,n ∈ 0} is σ(F , F )-dense in F . ′ (iv) This follows easily from the definitions, since (xn,xn)n∈N0 is a biorthogonal system. ✷ Lemma 2.1 is a version for separable Fr´echet spaces with a continuous norm of [6, Lemma 2], which was relevant in linear dynamics. See also [1, Lemma 2.11]. Observe that ′ ′ the existence of a sequence (en)n∈N0 ⊂ F satisfying (ii) and (iii) in Lemma 2.1 implies that the space F has a continuous norm. In fact, by (ii) there is a seminorm p on F such 2 ′ that supn∈N0 |en(x)| ≤ p(x) for all x ∈ F . Suppose that p(y) = 0 for some y ∈ F . Then ′ N en(y) = 0 for each n ∈ 0. Condition (iii) implies that y = 0. The statement and proof of our next result are inspired by Mashreghi and Ransford [11, Theorem 1.3]. Theorem 2.2 Let F be a separable, infinite dimensional, complex Fr´echet space with a continuous norm. Let G be an open connected domain in C. Then there exists a Fr´echet space E ⊂ H(G) of holomorphic functions on G such that F is isomorphic to E and the polynomials are dense in E. Proof. We apply Lemma 2.1 to the space F to select the sequences (en)n∈N0 ⊂ F and ′ ′ (en)n∈N0 ⊂ F satisfying conditions (i)−(iv). In particular, we find a continuous norm p on ′ N F such that |en(x)|≤ p(x) for each x ∈ F and each n ∈ 0. Now take a sequence (αn)n∈N0 ∞ n of positive numbers such that Pn=0 αnk < ∞ for each k ≥ 1. Define T : F → H(G) ∞ ′ n by T (x)(z) := Pn=0 αnen(x)z for each x ∈ F and each z ∈ G. The operator T is well- defined, linear and continuous. First of all, the series defining T (x) converges and defines an entire function for each x ∈ F , since ∞ ∞ ′ n n X αn|en(x)||z| ≤ p(x) X αnk < ∞ n=0 n=0 for each |z| ≤ k and each k ≥ 1. The continuity of T can be seen as follows: given an arbitrary compact subset L of G, there is k ≥ 1 such that |z| ≤ k for each z ∈ L. Then, for each x ∈ F we get ∞ ∞ ′ n n sup |T (x)(z)| = sup X αnen(x)z ≤ X αnk p(x). ∈ ∈ z L z L n=0 n=0 ′ Moreover, the map T is injective. Indeed, if T (x)=0in H(G), then en(x) = 0 for each ′ ′ ′ n ∈ N0. Since span({en | n ∈ N0}) is dense in (F ,σ(F , F )), this implies that x = 0. −1 n The monomials are clearly contained in T (F ), because T ((αn) en) = z for each N s n s −1 n ∈ 0.
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